Question about dimensional analysis

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Homework Help Overview

The discussion revolves around dimensional analysis in the context of physics equations related to motion and force. Participants are examining the units of various constants in equations involving motion and force, specifically focusing on the variable k.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to determine the unit of k in the equation for motion, questioning whether it could be 1/seconds^{1.5}, 1/seconds, or something else entirely. Another participant confirms one of the options. The original poster also seeks clarification on the unit of k in a force equation, leading to further validation from another participant. Additionally, the original poster expresses a desire for more practice with graphical analysis of equations, indicating uncertainty in their understanding of the topic.

Discussion Status

The discussion is active, with participants providing confirmations and clarifications regarding the units of k in different contexts. There is an ongoing exploration of graphical analysis techniques, with the original poster seeking additional resources for practice. While some guidance has been offered, there is no explicit consensus on the best resources or methods for practice.

Contextual Notes

The original poster mentions a lack of proper instruction in physics, which may contribute to their uncertainty and desire for additional practice problems. They also reference specific equations and their attempts to fit them into linear forms for analysis.

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Homework Statement



I have an equation for motion:
x(t) = x(0) + x(0) * k * t^{1.5}
x is in meters and t is in seconds. I have to determine the unit of k.

Homework Equations





The Attempt at a Solution



x(t) = x(0) + x(0) * k * t^{1.5}
[x(t)] = meters
Therefore, [x(0)] = meters and [x(0) * k * t^{1.5}] = meters
meters * [k] * seconds^{1.5} = meters
meters * (1/seconds^{1.5}) * seconds^{1.5} = meters

Does this mean the the unit of k is 1/seconds^{1.5}, 1/seconds, or something completely different?
 
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You first option is correct 1/s^1.5
 
Thanks, I didn't know if that was possible. I have another question if you could please check:

A force F is equal to k*x^n, where x is in centimeters.

[k*x^n] = N
[k]*(cm^n) = N
[k] = N/(cm^n)

The unit of k is N/(cm^n), is this correct?
 
Last edited:
Yes, you seem to have it sorted out!
 
Great, thanks. I've done far better in more difficult subjects like multivariable calculus, but the lack of a proper instructor for physics has me making stupid mistakes these days. :frown:

Would you happen to know of any online resource where I can practice graphical analysis of equations like these? What I have to do is to take a non-linear data set and convert it to a straight-line equation, determine appropriate units for slope and intercept, and determine values for the constants based on slope and intercepts. I only have two practice problems to work with.
 
I have never looked for any online resources for these types of problems.
If you know the power law for the equation (t^1.5 in your first example, n in your second example) Then the graph to plot is x against t^1.5 for the first and F against x^n for the second.
These would give straight lines with gradient k in each case.
If you do not know the power law... I think that is the case in your second example, you only know it as n then you must take logs :
F = k * x^n
LnF = Lnk + n*Lnx

A graph of LnF against Lnx will be a straight line with gradient n and intercept Lnk from which k can be calculated

Does this make any sense for you, have you met log ~ log graphs
 
I get the hang of those graphs, but I would feel more comfortable with practice. The other problem I have which I can't work is finding the resistivity p of a wire whose resistance R = (4pL)/(pi*d^2)

Plot R on the y-axis and 1/(d^2) on the x-axis to get gradient = 4pL/pi, and then p = gradient/(4L/pi). But I didn't get a correct graph:
https://www.physicsforums.com/showpost.php?p=3618742&postcount=3

Hence, that is why I would feel more comfortable with more to practice with. But I understand what you said about log ~ log graphs, I have this problem (the one I made this thread for, not the previous one) fitted into a linear form properly, all the points rest on a straight line.
 

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